Spherically symmetric solutions and the fundamental solution of biharmonic equation in $\mathbb R^n$

Question

Let $u:\;\mathbb R^n \rightarrow \mathbb R$

1. Find all the spherically symmetric solutions for the $\mathsf Δ^2 u=0$
2. Find one fundamental solution of the above equation

1) To begin with, I thought to turn the biharmonic operator in spherical coordinates and since we're talking about spherically symmetric solutions I wrote:

$(\frac{\partial ^2}{\partial r^2}\;+\;\frac{n-1}{r}\frac{\partial}{\partial r})(\frac{\partial ^2 v}{\partial r^2}\;+\;\frac{n-1}{r}\frac{\partial v}{\partial r})=0\;\;for\;u(x)=v(r)\;where\;r=\vert x \vert\;\;and\;x\in \mathbb R^n\;\;$

and then by computation I concluded to this:

$v_{rrrr}+\;\frac{2(n-1)}{r} v_{rrr}+\;\frac{(n-1)^2}{r^2} v_{rr}+\;\frac{(n-1)(3-n)}{r^3} v_{r}=0\;$. Then, I set $w(r)=v_{r} (r)\;$ and so the previous equation became $w_{rrr}+\;\frac{2(n-1)}{r} w_{rr}+\;\frac{(n-1)^2}{r^2} w_{r}+\;\frac{(n-1)(3-n)}{r^3} w=0\;$

Now, this is Euler's equation. I searched for solutions like $r^λ\;$ but at this point I got stuck! I find this : $ λ^3+(2n-5)λ^2+(n^2-4n+5)λ+(n-1)(3-n)=0\;$ which is something I cannot solve.

2) It holds that any solution of Laplace equation is also a solution of biharmonic equation. Is it ok for somebody to claim any fundamental solution of Laplace equation is also a fundamental solution of biharmonic equation?

Do I miss something? Are there any wrong computations? Hints for other solutions than this are welcome of course.

Answer 1

Generally, when an ODE is factored as $L_1L_2v=0$, you should take advantage of the factored form, not to multiply $L_1L_2$ out. It's easier to solve two ODEs of second order than one ODE of 4th order.

In this case $L_1=L_2=L$, where $Lv = r^{1-n}(r^{n-1}v')'$. The equation $Lv=0$ has general solution $Ar^{2-n}+B$ (with a logarithm appearing here when $n=2$, this case should be treated separately).

Then, $L(Lv)=0$ reduces to $Lv = Ar^{2-n}+B$. Using the above form of $L$ leads to $(r^{n-1}v')' = Ar + Br^{n-1}$, hence $r^{n-1}v' = Ar^2 + Br^{n} + C$ (not the same constants as before), $v' = Ar^{3-n} + Br + Cr^{1-n}$ and finally $$v = Ar^{4-n} + Br^2 + Cr^{2-n}+D$$ Again, the cases when one of the exponents in this computation is zero $(n=2,3,4)$ should be considered separately.

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For the fundamental solution, you want $\Delta u$ to be the fundamental solution of the Laplace equation, so that $\Delta^2 u$ is the delta function. This means $B=0$ above; the constants $C$ and $D$ won't be needed either. (But do pay attention to small dimensions: the answer is not $r^{4-n}$ in dimensions $\le 4$.)